An electronic assembly consists of two subsystems, say, $A$ and $B$ . From previous testing procedures, the following probabilities are assumed to be known :
$P(A{\rm{ fails }}) = 0.2$
$P(B{\rm{ fails alone }}) = 0.15$
$P(A{\rm{ and }}B{\rm{ fail }}) = 0.15$
Evaluate the following probabilities
(i) $P(A{\rm{ fails }}|B$ has failed)
(ii) $P(A{\rm{ fails a lone }})$
: Let $\bar A$ and $\bar B$ denote the events $A$ fails and $B$ fails respectively
Thus,
$P(A{\rm{ fails }}) = P(\bar A) = 0.2$
$P(A{\rm{ and }}B{\rm{ fail) }} = P(\bar A \cap \bar B) = 0.15$
$P(B{\rm{ fails alone }}) = P(\bar B) - P(\bar A \cap \bar B) = 0.15$
$\Rightarrow P(\bar B) - 0.15 = 0.15 \Rightarrow P(\bar B) = 0.30$
(i) $P(\bar A|\bar B) = \cfrac{{P(\bar A \cap \bar B)}}{{P(\bar B)}} = \cfrac{{0.15}}{{0.30}} = \cfrac{1}{2} = 0.5$
(ii) $P(A{\rm{ fail alone }}) = P(\bar A) - P(\bar A \cap \bar B) = 0.2 - 0.15 = 0.05$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Probability. Curated by Sachin Sharma. Free for all students.